3. The physics of colliding and exploding objects
The esoterica of colliding bodies can be difficult to grasp, but the basic principles need not be. This page explains the two basic principles and one or two of the corollaries most relevant to understanding JFK's motions as displayed in the Zapruder film after he was hit in the head from the rear.
There are two universal principles of the physics of colliding bodies, first that momentum in any directions is conserved, and second that total energy is conserved. The conservation of momentum is much easier to apply than the conservation of energy, for it is the sum of potential and kinetic energy that is conserved, and potential energy is often difficult or impossible to observe directly.
Momentum is defied as the product of mass and velocity (mv), and is often given the symbol p (p = mv). Momentum is a vector because velocity is a vector. That is why we speak of momentum being conserved in a given direction (actually in every direction). Energy is a scalar quantity (directionless). For example, kinetic energy contains a v2 term (KE = mv2/2), and the square of a velocity cannot be a vector.
In other words, momentum is conserved in every collision, whereas kinetic energy is not always. (In fact, it is seldom conserved.) The special types of collisions that conserve kinetic energy are called elastic collisions. Collisions that do not conserve kinetic energy are called inelastic. Because momentum is always conserved, it is easier to use to begin to understand collisions. (We will follow this procedure here.) But the lack of conservation of kinetic energy is very important to understanding the explosion of JFK's head, and we will discuss it as well.
Conservation means simply that the value of a quantity (momentum of energy here) is conserved during the collision, that is, that its value after the collision is the same as its value before. Conservation holds for the system of bodies, not for any individual body. Thus the sum of the momentums before the collision equals the sum after, and the same for energy.
We can express these conservation laws in simple conservation equations. Imagine two colliding bodies signified by the subscripts 1 and 2. For any given direction, their conservation of momentum appears as:
m1v1 + m2v2 = m1v1' + m2v2'
where the prime refers to conditions after the collision. We can generalize this equation for any number of bodies by using the Greek Σ to designate the sum over all the bodies:
Σ mivi = Σ mi'vi'
where i = 1, 2, 3, etc. For the usual x, y, and z axes, we can write:
Σ pxi = Σ pxi'
Σ pyi = Σ pyi'
Σ pzi = Σ pzi'
(But the axes don't have to be perpendicular in order that momentum be conserved. It is conserved in any direction and in all directions simultaneously.)
For the conservation of energy, we can write:
Ebefore = Eafter
which is the same as
KEbefore + PEbefore = KEafter + PEafter
The notation with primes is simpler:
KE + PE = KE' + PE'
For a collection of particles, this becomes
Σ KEi + Σ PEi = Σ KEi' + Σ PEi'
Σ mivi2/2 + Σ PEi = Σ mivi'2/2 + Σ PEi'
Examples of conservation of momentum
Let us first consider some qualitative implications of the conservation of momentum. The simplest system of colliding particles is two particles. Imagine one at rest and the other approaching from the left (with a positive x-velocity). Let us say that the particles collide head-on, and that the second particle moves off straight to the right. How will the first particle be moving? The first thing we can say is that it will not be moving in the y or z direction, because the first particle had no momentum in either of those directions and the second particle also did not (after the collision). Conversely, if the second particle moved off in the y direction (say upward), the first particle would have to take on a velocity in the negative y direction (downward) in order to make the net y-momentum after the collision sum to zero (its value before the collision).
Let us now consider an actual example. Suppose the masses m1 and m2 are 2 lb and 1 lb, respectively (we use English units here because the later calculations use them), and the initial velocities in the x-direction are 2 feet per second (fps) and 0 fps, respectively. Suppose also that the first body has a final x-velocity of 1.5 fps after the collision and no y-velocity. What will the final x-velocity of the second body be? To find the answer, we just set up the conservation equation for x-momentum and solve algebraically for the missing value v2x':
m1v1x + m2v2x = m1v1x' + m2v2x'
v2x' = (m1v1x + m2v2x - m1v1x')/m2
= (2x2 + 0 - 2x1.5) /1 = 1 fps
What will be the y-velocity of the second body? Do the same thing, but with y instead of x:
m1v1y + m2v2y = m1v1y' + m2v2y'
v2y' = (m1v1y + m2v2y - m1v1y')/m2
= (2x0 + 0 - 2x0) /1 = 0 fps
But we knew that ahead of time because of the reasoning given above.
Now consider what happened to the kinetic energy in this collision. We won't worry about the units of energy here, because they are too tricky in the English system. We will just determine whether the kinetic energy was conserved.
The initial KE was (2x22)/2 + 0 = 4 "energy units." The final KE was (2x1.52)/2 + (1x12)/2 = 2.75 energy units. Thus, about 30% of the kinetic energy was lost, making this an inelastic collision. Where did that 30% of the KE go? The conventional answer is that is was degraded into heat, but we shall see below that it can sometimes wind up in other forms.
Transformation of energy in extremely inelastic collisions
Of interest of us is the special case of a fast-moving light object striking a much heavier object and sticking to it. (This resembles the bullet hitting JFK's head.) Imagine, for example, a 1-lb projectile moving at 2000 fps colliding with a 100-lb stationary object and sticking to it. We address two questions: What is the final velocity of the pair, and what happens to the kinetic energy in the process?
To find the final velocity of the pair, we use the same equation for conservation of momentum that we used above, but with the same final velocity for both objects, which we will call v1'.
m1v1 + m2v2 = m1v1' + m2v1' = (m1 + m2)v1'
v1' = (m1v1 + m2v2)/(m1 + m2)
= (1x2000 + 100x0)/(1 + 100) = 19.8 fps
That's the straightforward part. The kinetic energy part
is much more interesting. The initial KE is 1x20002/2 + 0 = 2,000,000
"energy units." The final KE is 101x19.82/2 = 19,798 units,
or only 1% of the original. In other words, 99% of the KE has disappeared in
this highly inelastic collision. Where did that 99% go? It disappeared into
deformation and something like heat, warming the deformed and fused bodies after they collided.
This very significant aspect of inelastic collisions is seldom appreciated, but it is vital to understanding the JFK assassination. It means that when a bullet hit the head, the great majority of the initial KE will disappear into the head, where it will temporarily be kept in reserve. If the head does not explode, the energy will be degraded into heat and never seen or heard from again. But if the head explodes, the KE can be released to work again, this time in the form of explosive energy of fragments and of the head and body in recoil to those fragments. The KE available this way is much greater than any KE given to the body by the direct impact of the bullet. That is why JFK's body can potentially recoil in any direction if his head explodes. Think of the explosion as a way to release all that pent-up energy from the hit.
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